If you flip a fair coin $7$ times, what is the probability that you will get exactly $4$ tails?
One way to solve this problem is to figure out how many ways you can get exactly $4$ tails, then divide this by the total number of outcomes you could have gotten. Since every outcome has equal probability, this will be the probability that you will get exactly $4$ tails. How many outcomes are there where you get exactly $4$ tails? Try thinking of each outcome as a $7$ -letter word, where the first letter is "H" if the first coin toss was heads and "T" if it was tails, and so on. So, the number of outcomes with exactly $4$ tails is the same as the number of these words which have $4$ T's and $3$ H's. How many of these are there? If we treat all the letters as unique, we'll find that there are $7!$ different arrangements, overcounting $4!$ times for every time we only switch the T's around, and $3!$ times for every time we only switch the H's around. So, there are $\dfrac{7!}{4!3!} = 35$ different outcomes where you get exactly $4$ tails. Altogether, there are $2^{7} = 128$ total possible outcomes. So, the probability that you will get exactly 4 tails is $\dfrac{35}{128}$.